$A$ wheel with a moment of inertia of $5 \times 10^{-3} \ kg \cdot m^2$ is rotating at a rate of $20 \ rev/s$. The torque required to bring it to rest in $10 \ s$ is:

Find the ratio of linear acceleration $(a)$ to angular acceleration $(\alpha)$ of the center of mass $(COM)$ for the given diagram. Given: $m = 2 \, kg$,$r = 10 \, cm = 0.1 \, m$,and force $F = 20 \, N$ applied at the center.

$A$ torque of $10^5 \ N \cdot m$ is applied to a disc of mass $10 \ kg$ and radius of gyration $50 \ m$. What is the value of the angular acceleration in $rad/s^2$?

$A$ rod $PQ$ of mass $M$ and length $L$ is hinged at end $P$. The rod is kept horizontal by a massless string tied to point $Q$ as shown in the figure. When the string is cut,the initial angular acceleration of the rod is

$A$ wheel having a moment of inertia $2 \; kg \cdot m^2$ about its vertical axis rotates at the rate of $60 \; rpm$ about this axis. The torque required to stop the wheel's rotation in one minute is:

$A$ wheel of radius $0.4 \,m$ can rotate freely about its axis as shown in the figure. $A$ string is wrapped over its rim and a mass of $4 \,kg$ is hung. An angular acceleration of $8 \,rad \,s^{-2}$ is produced in it due to the torque. Then, the moment of inertia of the wheel is $(g = 10 \,m \,s^{-2})$.